Setanta, 88 bytes

gniomh(s){a:=0le i idir(0,fad@s){i=s[i]a+=(i=='['|0&1)-(i==']'|0&1)ma a<0 bris}toradh!a}

Try on try-setanta.ie

Uiua, 23 bytes

⍤./↧=0⊂⊢⇌:<0.\+-∩=@],@[

Try it!

JCram, 13 bytes.

5,⍥ψ⍺⍘ⒼιV⍣b⍐B

Encodes the ES6 program

f=s=>(t=s.replace(/\[\]|[^\[\]]/g,''))==s?!s:f(t)

Vyxal, 10 bytes

‛[]↔øβ¬[°*

Try it Online!

The joys of having a balanced-bracket digraph.

Explained

‛[]↔øβ¬[°*
‛[]↔       # remove all non "[]" from the input
    øβ¬    # are the brackets unbalanced?
       [°* # if so, multiply the input by the imaginary number `i`. Obviously you can't multiply a string by an imaginary number in python, so this errors.

Turing Machine Code, 286 272 bytes

Again, I'm using the rule table syntax defined here.

0 * * l 1
1 _ ! r Q
5 _ _ r 5
5 * * * W
W [ ! l E
W ] ! l T
W _ _ l I
W * ! r *
E ! ! l *
E * * * R
R _ [ r O
R * * l *
T ! ! l *
T * * * Y
Y _ _ r U
Y * * l *
U [ _ r O
U ! ! * halt-r
I ! ! l *
I _ _ r halt
I * * l halt-r
O ! ! r Q
O _ _ l I
O * * r *
Q ! ! r *
Q * * * W

Terminates in state halt to accept the input and halt-r to reject it.

Stax, 14 11 characters

╬Äτ↔EªÉs «Ü

Run and debug it

Credit to @recursive for -3 bytes.

ASCII equivalent:

.[]Y|&{yz|egp!

Remove all characters except [], then remove [] until the string no longer changes. Return 1 if the final string is empty.

Haskell (167,159)

Did this mostly for fun, if anyone's got any suggestions for how to make it shorter, I'd be glad to hear them tho :)

import System.Exit
p x y b|b=x|True=y
main=getContents>>=(return.all (>=0).scanl (\v->p (v-1) (v+1).(==']')) 0.filter (`elem`"[]"))>>=p exitSuccess exitFailure

Edit: Fixed an issue pointed out to me in the comments (added 11 bytes).

Edit 2: Created auxiliary function for testing predicates using guards inspired by user13350, removing 8 bytes.

Pyth, 25 bytes

Ilzv::z"[^[\]]"k"]\[""],[

Try it online!

import re
z=input()
if len(z):
 print(eval(re.sub("]\[","],[",re.sub("[^[\]]","",z))))

Hassium, 104 Bytes

func main(){l=0;r=0;b=input();foreach (c in b){if(c=="[")l++;if(c=="]")r++;}if(!(r==l))exit(1);exit(0);}

Full expanded (note doesn't work in online interpreter as input() is disabled) here

Ruby, 59 58

Scans for opening and closing bracket, counting [ as 1 and ] as -1, and exits if the count drops below 0.

b=0
$<.read.scan(/\]|\[/){exit 1if(b+=92-$&.ord)<0}
exit b